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Direct Common Tangent

In a right an...

Question

In a right angled triangle, circumradius R is equal to

\frac{s + r}{2}

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\frac{s - r}{2}

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s - r

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\frac{s + r}{a}

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Solution

The correct option is B $\frac{s - r}{2}$
[Considering co-ordinate axes].
$R X^{2}$ = $B X^{2}$ = $\frac{1}{4} (a^{2} + c^{2}) = \frac{1}{4} b^{2}$ (Pythagoras theorem)
$R = \frac{b}{2}$
And, $r = (s - b) tan \frac{B}{2}$
r=(s-b)
R= $\frac{s - (s - b)}{2}$
$R = \frac{s - r}{2}$
Hence, $\frac{s - r}{2}$ is the correct answer,

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